A container has an unknown quantity of gas at a pressure of 121.6kPa, a volume of 31L and a temperature of 87oC. How many moles of gas are present ?

1.26mol

Explanation:

Using the general gas law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K).

Based on the information provided in this question,

P = 121.6kPa = 121.6 × 0.00987atm = 1.2001 atm

V = 31L

T = 87°C = 87 + 273 = 360K

R = 0.0821 Latm/molK

Using PV = nRT

n = PV/RT

n = (1.2001 × 31) ÷ (0.0821 × 360)

n = 37.2031 ÷ 29.556

n = 1.258

n = 1.26mol

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1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%